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# I know this one has been posted before by Kevincan but i

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I know this one has been posted before by Kevincan but i [#permalink]

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10 Aug 2006, 11:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I know this one has been posted before by Kevincan but i read the xplanation over and over and still cant get the sense of it.

I will highly appreciate if someone can explain it to me with Just logic and not rocket science math.

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.

Thanks
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10 Aug 2006, 11:38
D? whats the OA?

The only situation where the average of the squares will be less then the original average (given that all are +), is when all 9 numbers are 0<x<1.

S1 - means that there is atleast 1 number thats greater then 1.
S2 - means, again, that there is atleast 1 number greater then 1.

Does this make sense, or am I talking total nonsense? I haven't read the preivous post on this.
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10 Aug 2006, 11:57
OA is B foe some weired reason.

Here you are another one that i didn't get the explanation to it too.

Please if you understands the explanation give me a ring

S is a set of negative integers. T is a subset of S, and there are two fewer elements in T than in S. Is the mean of the elements in S different from that of the elements in T?

(1) The mean of the elements in S is between -4.5 and -5.0.
(2) At least three elements in S are single-digit negative integers.

IMO, A

S1. -5.0<mean<-4.5 tells that the mean can't be the same.

Set S = {...........,X1,X2}
Set T = {...............}
Let mean of set S = s and sum of set S = T1
Let mean of set T = t and sum of set t = T2

If mean of S is equal to mean of T
T1/n = (T1-X1-X2)/n-2

T1 = (X1+X2)n/2 (n is number of elements in set S)
T2 = (X1+X2)n/2-X1-X2 = (X1+x2)(n-2)/2
mean of set S : T1/n = (X1+X2)/2
mean of set T : T2/n-2 = (X1+X2)/2
since X1 and X2 are integers, (X1+X2)/2 is either integer value or .5
but S1. -5.0<mean<-4.5 thus it contradicts assumption. i.e. the mean of set S can't be the same as the mean of set T
Thus, sufficient.

S2. At least three elements in S are single-digit negative integers.
Ex1.
set S = {-1,-2,-3} set T = {-2}; -2 = -2 :
but
set S = {-1,-2,-3} set T = {-1}; -2!=-1 :
Thus insufficient.
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10 Aug 2006, 12:21

1.) this means that at least one of the numbers is greater than 1, and that whenever it is squared, it will indefinitly increase the sum of the 9 numbers

2.) Same for this one. Assuming that it is the lowest numbers possible, there would still be at least a 2. When 2 is square, it would also increase the sum of the 9 numbers.
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10 Aug 2006, 12:30
Here is the explanation that was posted ( i see it rocket science)
Those folks know things i ve never heared about in my entire life

Proof of stmt B.

difference between Y 9th term and X 9th term

1. Y9-X9 > 2

let
F(x) = x - x^2
F(x) = -(x-1/2)^2 + 1/4 (parabola, concave down)
Thus, when x = 1/2 we got maximun value of F(x) = 1/4

Assuming that sum of differences between each term of Set X and each term of Set Y, 1st through 8th term, has the maximun value
2. (1/4)*8 = 2

Y9-X9 is always greater than 2. (Y9-X9 >2 and (X1-Y1)+(X2-Y2)..+(X8-Y8) <= 2)

Sufficient.
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10 Aug 2006, 12:42
Hi rdw28,

You have a point but if we assumed that the 4 numbers( memebers of the oriinal set are fractions .... ie by squaring they will exponantially decrease because originally they can be expressed in this way

ie: 1l8 = 2^-3 so if we squared this fraction given that the 4 memebers of the set after the median (1).are just a repeated ( 1.0000001) it ll never compensate for the decrease in the sum of the set.

And regarding the second statment. if the range is 2 ie the elements of the set are between ( x , x+2) representing the 1st and 9th members of the set respectively if the set is rranged ascendingly.

if the first term is 1/8 therfore the largest will be 2 1/8 thus by squaring it will give a bigger value because the two sides ( before and after the mean are increasing exponantially " squared)

if the first term is -1 therfore the largest is 3.........same as the first scenario.

This is my own understanding to the problem but ......my problem is the explanation of the OA posted because my way is time consuming .

if the first term
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10 Aug 2006, 12:51
Hey yezz,

Thanks alot. I understand your explanation. I just told myself that the increase in the whole integer would offset that average totally; however, my mistake nonetheless.

In regards to B - I guess once you know the range of the highest and lowest numbers, you can figure whether or not if it is greater.
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Re: Hardest mean - average to understand ( Help needed) [#permalink]

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10 Aug 2006, 13:01
yezz wrote:
I know this one has been posted before by Kevincan but i read the xplanation over and over and still cant get the sense of it.

I will highly appreciate if someone can explain it to me with Just logic and not rocket science math.

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.

Thanks

St1: Lets take teh worst case scenrio
X = {a,a,a,a,b,b,b,b,b}
Y = {c,c,c,c,d,d,d,d,d}
where b is little greater than 1 and a is little less than 1.
Say if a = 0.5 and b = 1.01 c = 0.25 and d = 1.0201
Then increase in sum of Y is 0.1005 but decrease is 1.0 so sum of elements of Y is less. In case last five elements of X are much larger than 1 then sum of elements of Y will be greater. Hence INSUFF

St2: This means one element of X is greater than 2. If first four elements of X are less than 1 then due to these numbers sum of elements of Y can be reduced by 1 point only which will be compensated by the increase in the sum due to elements greater than 2.: SUFF
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10 Aug 2006, 13:53
Dahiya, I perceive you are a lady.

Thanks a million if i were to choose an instructor to guide me with GMAT , IT WOULD BE YOU .

i ve been going through many of the problems you solved.

and believe me when i tell you , you are the simplest logical analysis i have seen attacking most of the problems and if you d allow me .... we think on the same wave lenght.

Thanks a million once again.
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10 Aug 2006, 14:11
yezz wrote:
Dahiya, I perceive you are a lady.

Thanks a million if i were to choose an instructor to guide me with GMAT , IT WOULD BE YOU .

i ve been going through many of the problems you solved.

and believe me when i tell you , you are the simplest logical analysis i have seen attacking most of the problems and if you d allow me .... we think on the same wave lenght.

Thanks a million once again.

Learning from wonderful people at this forum. Thanks a lot for such nice words. (Actually I rarely receive such comments )

I am a full grown 5' 8" male weighing 165 pounds and also a high school time boxer

Want to see some of my photos? See this
http://www.gmatclub.com/phpbb/viewtopic.php?t=32623
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Re: Hardest mean - average to understand ( Help needed) [#permalink]

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10 Aug 2006, 14:13
ps_dahiya wrote:
St2: This means one element of X is greater than 2. If first four elements of X are less than 1 then due to these numbers sum of elements of Y can be reduced by 1 point only which will be compensated by the increase in the sum due to elements greater than 2.: SUFF

What if 8 numbers are less than 1?
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Re: Hardest mean - average to understand ( Help needed) [#permalink]

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10 Aug 2006, 14:20
kevincan wrote:
ps_dahiya wrote:
St2: This means one element of X is greater than 2. If first four elements of X are less than 1 then due to these numbers sum of elements of Y can be reduced by 1 point only which will be compensated by the increase in the sum due to elements greater than 2.: SUFF

What if 8 numbers are less than 1?

You are right. But then the max decrease these number can cause is 2. Still Y will be greater. Right???
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Re: Hardest mean - average to understand ( Help needed) [#permalink]

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10 Aug 2006, 14:25
hmm..

st1 says median of X is greater than median of Y..

ur example...X median is b, which is 1.01 while the median of Y is 1.02...kind of defies wht st1 says...

ps_dahiya wrote:

St1: Lets take teh worst case scenrio
X = {a,a,a,a,b,b,b,b,b}
Y = {c,c,c,c,d,d,d,d,d}
where b is little greater than 1 and a is little less than 1.
Say if a = 0.5 and b = 1.01 c = 0.25 and d = 1.0201
Then increase in sum of Y is 0.1005 but decrease is 1.0 so sum of elements of Y is less. In case last five elements of X are much larger than 1 then sum of elements of Y will be greater. Hence INSUFF

St2: This means one element of X is greater than 2. If first four elements of X are less than 1 then due to these numbers sum of elements of Y can be reduced by 1 point only which will be compensated by the increase in the sum due to elements greater than 2.: SUFF
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Re: Hardest mean - average to understand ( Help needed) [#permalink]

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10 Aug 2006, 14:31
ps_dahiya wrote:
kevincan wrote:
ps_dahiya wrote:
St2: This means one element of X is greater than 2. If first four elements of X are less than 1 then due to these numbers sum of elements of Y can be reduced by 1 point only which will be compensated by the increase in the sum due to elements greater than 2.: SUFF

What if 8 numbers are less than 1?

You are right. But then the max decrease these number can cause is 2. Still Y will be greater. Right???

Right Your explanation would have been perfect if you had said that. Great work
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10 Aug 2006, 14:32
Sorry if i offended you (though being a lady is a wonderfull thing to be )......We dont want ladies to make a strick or something.

Still i stick to the same openion about your logical reasoning to maths.

By the way ( great adventurous photos)
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Re: Hardest mean - average to understand ( Help needed) [#permalink]

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10 Aug 2006, 21:28
kevincan wrote:
Right Your explanation would have been perfect if you had said that. Great work

Kevin,

You are a true teacher. Thanks for pointing such silly mistakes. Though I don't like Kaplan Tests but I like you
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Re: Hardest mean - average to understand ( Help needed)   [#permalink] 10 Aug 2006, 21:28
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